Explain the Pattern - Part 3

]Just choose the numbers so that $\overline{def}$ is the reverse of $\overline{abc}$ . Then $\overline{fed} = \overline{abc}$ and $\overline{cba} = \overline{def}$ , and so the identity $\overline{abc} \times \overline{def} = \overline{fed} \times \overline{cba}$ is trivial. The trick is then to make this common product a palindrome. Since $\overline{abc} \times \overline{cba} \; = \; 10000ac + 1000(a+c)b + 100(a^2+b^2+c^2) + 10(a+c)b + ac$ this can be done easily by ensuring that $ac$ , $(a+c)b$ and $a^2+b^2+c^2$ are single digit numbers. Thus $\{a,c\} = \{1,2\}$ (since $a,c$ must be nonzero and distinct), and $b=0,1,2$ . We have the solutions $102 \times 201 = 20502 \hspace{2cm} 112 \times 211 = 23632 \hspace{2cm} 122 \times 221 = 16962$

The same approach will provided the $4$ -digit solution $1112 \times 2111 = 2347432$ , with a $4$ digit number, times its reverse, equalling a palindrome. We can find a $5$ digit solution as well: $11012 \times 21011 = 231373132$ . Indeed, we can find an example of any length: $1\underbrace{00...0}_{n}2 \times 2\underbrace{00...0}_{n}1 \; = \; 2\underbrace{00...0}_{n}5\underbrace{00...0}_{n}2$

for any nonnegative integer $n$ .

$122 \times 213 = 25986$ and $68952 = 312 \times 221$ .

I do not know how to find this other than trial and error. The numbers seem similar to the 2 digit case, but I don't think it's special.

Does a 4-digit pattern also exist?

I multiplied out (100a + 10b + c)(100d + 10e + f) and then aimed for no carrying into next columns. This lowered the amount of trial and error, since:

ae+bd<10 (for the 1000's column)

af+be+cd<10 (for the hundreds column)

And that was enough to find

123 x 112 = 13776 while 67731 = 211 x 321

I like Mark Hennings solution with ghijk being a palindrome, because it generalises.